JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Haloalkanes and Haloarenes

A Grignard reagent formed from $\mathrm{RBr}$ and $\mathrm{Mg}$ in dry ether, then quenched with water, simply replaces the C–Br bond with a C–H bond — the carbon skeleton of $\mathrm{R}$ carries through unchanged into the alkane $\mathrm{R}-\mathrm{H}$.

So the question reduces to: how many structural isomers of ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{Br}$ share the carbon skeleton of 2-methylbutane, $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CHC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$?

Placing $\mathrm{Br}$ on each distinct carbon of that skeleton gives:
1-bromo-2-methylbutane, $\mathrm{BrC}{\mathrm{H}}_2\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$
2-bromo-2-methylbutane, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}(\mathrm{Br})(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$
2-bromo-3-methylbutane, $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CHCH}(\mathrm{Br})\mathrm{C}{\mathrm{H}}_3$
1-bromo-3-methylbutane, $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CHC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$

Each of these four bromides carries the 2-methylbutane skeleton, so Grignard formation followed by hydrolysis regenerates 2-methylbutane in every case.

Four structural isomers of $\mathrm{RBr}$ give 2-methylbutane by this route.